Coordinate Geometry Test

Result

Coordinate Geometry Test - 43

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Question 1
1 / -0

The points $$( - a , - b ) , ( 0,0 ) , ( a , b ) , \left( a ^ { 2 } , a b \right)$$ are

A
collinear

B
vertices of a parallelogram

C
concyclic

D
vertices of a rectangle

Solution
Question 2
1 / -0

If $$P \left( \dfrac{a}{3},\dfrac{b}{2} \right)$$ is the mid-point of the line segment joining $$A(-4,3)$$ and $$B(-2,4)$$, then $$(a,b)$$ is

A
$$(-9,7)$$

B
$$\left( -3, \dfrac{7}{2} \right)$$

C
$$(9,-7)$$

D
$$\left( 3, -\dfrac{7}{2} \right)$$

Solution

Using the midpoint formula, $$\left(x,y\right)=\left(\dfrac{{x}_{1}+{x}_{2}}{2},\dfrac{{y}_{1}+{y}_{2}}{2}\right)$$ where $${x}_{1}=-4$$,$${y}_{1}=3$$,$${x}_{2}=-2$$, $${y}_{2}=4$$
Given $$P$$ is a mid-point of $$AB$$
$$\Rightarrow P\left(\dfrac{a}{3},\dfrac{b}{2}\right)=\left(\dfrac{-4-2}{2},\dfrac{3+4}{2}\right)=\left(-3,\dfrac{7}{2}\right)$$
Equating the $$x$$ and $$y$$ coordinates, we get
$$\Rightarrow \dfrac{a}{3}=-3,\dfrac{b}{2}=\dfrac{7}{2}$$
$$\Rightarrow a=-9,b=7$$
$$\therefore \left(a,b\right)=\left(-9,7\right)$$
Question 3
1 / -0

The line segment joining the points $$(1, 2)$$ and $$(k, 1)$$ is divided by the line $$3x + 4y - 7 = 0$$ in the ratio $$4 : 9$$ then $$k$$ is

A
$$2$$

B
$$-2$$

C
$$3$$

D
$$-3$$

Solution
Question 4
1 / -0

If area of a triangle is $$35$$ square units with vertices $$\left( {2, - 6} \right),\,\,\left( {5,\,\,4} \right)$$ and $$({k},\,\,4)$$ then $${k}$$ is :

A
$$ -2 \space \text{or} \space 12$$

B
$$12$$

C
$$-2$$

D
$$ 2 \space \text{or} \space 12$$

Solution
Question 5
1 / -0

If the ratio in which the line segment joining the points (6,4) and (x,-7) divided internally by y-axis is 6: 1, then x equals

A
2

B
3

C
-1

D
-2

Solution
Question 6
1 / -0

If $$(-6,-4)$$ and $$(3,5)$$ are the extremities of the diagonals of a parallelogram and $$(-2,1)$$ is its third vertex, then its fourth vertex is

A
$$(-1,0)$$

B
$$(0,-1)$$

C
$$(-1,1)$$

D
none of these

Solution

Given,

$$P(-6,-4),Q(3,5),R(-2,1),S(\alpha ,\beta )$$

Let $$P$$ and $$Q$$ are the extremities of diagonals of a parallelogram, and

$$R$$ and $$S$$ will be the extremities of diagonals of a parallelogram

Now,

midpoint of $$PQ=\dfrac{3-6}{2},\dfrac{5-4}{2}=\dfrac{-3}{2},\dfrac{1}{2}$$

midpoint of $$RS\Rightarrow \dfrac{-2+\alpha }{2}=-\dfrac{3}{2}$$

$$\Rightarrow \alpha =-3+2=-1$$

Now,

$$\dfrac{1 +\beta }{2}=\dfrac{1}{2}$$

$$\Rightarrow \beta =0$$

Therefore, coordinates of 4th vertex is $$(-1,0)$$
Question 7
1 / -0

If $$P\left(\dfrac{a}{3},4\right)$$ is the mid point of the line segment joining the points $$(-6,5),(-2,3).$$ Then, find the value of $$a.$$

A
$$-4$$

B
$$-12$$

C
$$12$$

D
$$-6$$

Solution

Given: $$P(x,y)$$ is md-point of $$QR$$ then $$\left(\dfrac{a}{3}, 4\right) = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$$

$$\Rightarrow \left(\dfrac{a}{3}, 4\right) = \left(\dfrac{-6-2}{2}, \dfrac{5+3}{2}\right)$$

$$\Rightarrow \dfrac{a}{3} = \dfrac{-8}{2}$$

$$\Rightarrow a = -4 \times 3 = -12$$
Question 8
1 / -0

The mid-point of the line joining the points $$(2p +2, 3)$$ and $$(4, 2q + 1)$$ is $$(2p, 2q)$$, then

A
$$p = 3, q = 2$$

B
$$p = 2, q = 3$$

C
$$p = -3, q = 2$$

D
$$p = 2, q = 2$$

Solution
Question 9
1 / -0

If the mid point of $$(6, 8)$$ and $$(8, 6)$$ is mid point of $$(10, 3)$$ and $$(x, y)$$, then $$x + y$$ is

A
$$14$$

B
$$15$$

C
$$16$$

D
$$17$$

Solution
Question 10
1 / -0

The area of a triangle with vertices $$A(5,0),B(8,0)$$ and $$C(8,4)$$ in square units is

A
$$20$$

B
$$12$$

C
$$6$$

D
$$16$$

Solution

Area of Triangle using coordinates of A, B and C
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$$--------(1)
Here in our case $$ A(5,0) = (x_1, y_1)$$
$$ A(8,0) = (x_2, y_2)$$
$$ A(8,4) = (x_3, y_3)$$

Substituting above values in Eq - 1, we get
Area $$ = 0.5 [ 5(0-4) + 8(4-0) + 8(0-0)]$$
Area $$ = 0.5 [ -20 + 32]$$
Area $$ = 0.5 \times (12)=6$$
$$ Area = 6 $$ square units.